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A279125
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Lexicographically earliest sequence such that, for any distinct i and j, a(i)=a(j) implies (i AND j)=0 (where AND stands for the bitwise AND operator).
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5
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0, 0, 1, 0, 2, 3, 4, 0, 3, 2, 5, 1, 6, 7, 8, 0, 7, 6, 9, 5, 10, 11, 12, 4, 13, 14, 15, 16, 17, 18, 19, 0, 11, 10, 16, 9, 14, 13, 20, 12, 21, 22, 23, 24, 25, 26, 27, 1, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 0, 18, 17, 24, 15, 22, 21, 35, 9
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OFFSET
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1,5
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COMMENTS
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This sequence is similar to A279119 in the sense that here we check for common ones in binary representation and there we check for common prime factors.
By analogy with A275152, this sequence can be seen as a way to tile the first quadrant with fixed disconnected 2-dimensional polyominoes: the (vertical) polyomino corresponding to n is shifted to the right as little as possible so as not to overlap a previous polyomino, and a(n) gives the corresponding number of steps to the right (see illustration in Links section).
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LINKS
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FORMULA
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MAPLE
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with(Bits):
n:= 100:
l:= []:
g:=[seq(0, i = 0..n-1)]:
for i from 1 to n by 1
do
a:= 0;
while (And(g[a + 1], i)) > 0
do
a++;
end do:
g[a + 1] += i;
l:= [op(l), a];
end do:
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MATHEMATICA
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n = 100;
l = {};
g = ConstantArray[0, n];
For[i = 0, i < n, i++; a = 0; While[BitAnd[g[[a + 1]], i] > 0, a++];
g[[a + 1]] += i;
l = Append[l, a]];
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PROG
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(PARI) g = vector(72); for (n=1, #g, a = 0; while (bitand(g[a+1], n)>0, a++); g[a+1] += n; print1 (a", "))
(Python)
n = 100
g = n * [0]
for i in range(1, n + 1):
a = 0
while g[a] & i:
a += 1
g[a] += i
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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